Question

Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by the Institute for College Access and Success showed that in 69% of college graduates from public and nonprofit colleges in 2013 had student loan debt. A researcher wanted to see if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014. Suppose that the researcher surveyed 1500 graduates of public and nonprofit universities and found that 71% of graduates had student loan debt in 2014. Let p be the proportion of all graduates of public nonprofit universities that graduated with student loan debt. What are the appropriate null and alternative hypotheses for this research question?



A. H0: p = 0.69 Ha: p ≠ 0.69



B. H0: p = 0.69 Ha: p > 0.69



C. H0: p = 0.71 Ha: p ≠ 0.71



D. H0: μ = 0.69 Ha: μ > 0.69

Answer 1

Answer:

Null hypothesis:$p \leq 0.69$  

Alternative hypothesis:$p > 0.69$  

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

$\hat p=0.71$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.69$ is the value that we want to test

$\alpha$ represent the significance level  

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

Null hypothesis:$p \leq 0.69$  

Alternative hypothesis:$p > 0.69$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675$  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

$p_v =P(Z>1.675)=0.047$  

If we compare the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.